Question: Solve for $x$ and $y$ by deriving an expression for $x$ from the second equation, and substituting it back into the first equation. $\begin{align*}9x-3y &= -6 \\ -7x+6y &= 7\end{align*}$
Begin by moving the $y$ -term in the second equation to the right side of the equation. $-7x = -6y+7$ Divide both sides by $-7$ to isolate $x$ $x = {\dfrac{6}{7}y - 1}$ Substitute this expression for $x$ in the first equation. $9({\dfrac{6}{7}y - 1}) - 3y = -6$ $\dfrac{54}{7}y - 9 - 3y = -6$ Simplify by combining terms, then solve for $y$ $\dfrac{33}{7}y - 9 = -6$ $\dfrac{33}{7}y = 3$ $y = \dfrac{7}{11}$ Substitute $\dfrac{7}{11}$ for $y$ in the top equation. $9x-3( \dfrac{7}{11}) = -6$ $9x-\dfrac{21}{11} = -6$ $9x = -\dfrac{45}{11}$ $x = -\dfrac{5}{11}$ The solution is $\enspace x = -\dfrac{5}{11}, \enspace y = \dfrac{7}{11}$.